2

In the proof of "absolute convergence implies unconditional convergence" for a convergent series $\sum_{n=1}^{\infty}a_n$, we take a partial sum of first $n$ terms of both the original series ($S_n$) and rearranged series ($S_n'$) and compare them. Because the original series converges, we get some $N$ from Cauchy-criteria. Now if I choose "$n$ large enough" such that $\{a_1,a_2,\dots,a_{N-1}\} \subseteq \{a_1',a_2',\dots,a_{n}'\}$. Then if we compare both the partial sums. the remaining $a_i$s are all for $i \geq N$, but still some $a_i'$s are remaining. The book claims that $$|\sum_{i=N}^{n}a_i-\sum_{i=N}^{n}a_i'| \leq |\sum_{i=N}^{n}|a_i||.$$ I could not not understand how these $a_i'$s are getting removed.

An infinite series is called "unconditionally convergent" if every rearrangement of it converges to the same value. If the sequence is absolutely convergent, then it can be shown that all converges to the same value, in fact, the above theorem proves that.

Martin Sleziak
  • 53,687
user96000
  • 353
  • That's not right. It could be a misreading on your part, or a mistake in the book. It's $$\left\lvert\sum_{i=1}^n a_i - \sum_{i=1}^n a_i'\right\rvert \leqslant \sum_{i=N}^\infty \lvert a_i\rvert.$$ – Daniel Fischer Sep 21 '13 at 10:04
  • 1
    Duplicate maybe: http://math.stackexchange.com/questions/499568/absolute-convergence-implies-unconditional-convergence – Arash Sep 21 '13 at 10:18
  • @DanielFischer: sorry I typed it wrongly, corrected now. But, how the $a_i's$ are getting removed ? – user96000 Sep 21 '13 at 10:24
  • 1
    You wrote: The book claims. Which book is it? It would be nice to include reference. – Martin Sleziak May 22 '15 at 04:54

1 Answers1

3

If your text actually writes

$|\sum_{i=N}^{n}a_i-\sum_{i=N}^{n}a_i'| \leq |\sum_{i=N}^{n}|a_i||$

then it is indeed mistaken. (Note also that even on its own the right hand side is strangely written: why do we need the outside absolute value?) For instance, suppose $a_N = \ldots = a_n = 0$. Then the inequality implies $\sum_{i=N}^n a_i' = 0$, but the assumptions do not give us that.

I would suggest a somewhat different proof, namely the one which is given in Theorem 14.7 of these notes.

Put $A = \sum_{n=0}^{\infty} a_n$. Fix $\epsilon > 0$ and let $N_0 \in \mathbb{N}$ be such that $\sum_{n=N_0}^{\infty} |a_n| < \epsilon$. Let $M_0 \in \mathbb{N}$ be sufficiently large so that the terms $a_{\sigma(0)},\ldots,a_{\sigma(M_0)}$ include all the terms $a_0,\ldots,a_{N_0-1}$ (and possibly others). Then for all $M \geq M_0$, $$|\sum_{n=0}^{M} a_{\sigma(n)} - A| = |\sum_{n=0}^{M} a_{\sigma(n)} - \sum_{n=0}^{\infty} a_n| \leq \sum_{n=N_0}^{\infty} |a_n| < \epsilon.$$ Indeed: by our choice of $M$ we know that the terms $a_0,\ldots,a_{N_0-1}$ appear in both $\sum_{n=0}^M a_{\sigma(n)}$ and $\sum_{n=0}^{\infty} a_n$ and thus get cancelled; some further terms may or may not be cancelled, but by applying the triangle inequality and summing the absolute values we get an upper bound by assuming no further cancellation. This shows $\sum_{n=0}^{\infty} a_{\sigma(n)} = \lim_{M \rightarrow \infty} \sum_{n=0}^M a_{\sigma(n)} = A$.

Martin Sleziak
  • 53,687
Pete L. Clark
  • 97,892
  • @Peter L. Clark: "but by applying the triangle inequality and summing the absolute values we get an upper bound by assuming no further cancellation." Can you please elaborate this. It is not clear to me. – user96000 Sep 22 '13 at 11:14
  • 1
    The point is that if I add some terms of a series starting with $N_0$, then by the triangle inequality that is in absolute value less than or equal to the sum of the absolute values of those terms, which is less than or equal to the sum of the absolute values of every term indexed by $n \geq N_0$. – Pete L. Clark Sep 22 '13 at 13:51
  • @PeteL.Clark I think it should be $|\sum_{n=0}^{M} a_{\sigma(n)} - \sum_{n=0}^{\infty} a_n| \leq \sum_{n=N_0}^{\infty} |a_n| + \sum_{ 0\le n\le M, \sigma(n)\ge N_0}|a_{\sigma(n)}| < 2\epsilon$. The additional $\sum$ is for the terms which are not canceled. How could you remove that additional terms by triangle inequality ? – Hua Feb 18 '17 at 04:46
  • @Hua In my opinion, the inequation is not obtained by triangle inequality basically. When $\sum_{n=0}^\infty a_n$ minus $\sum_{n=0}^M a_{\sigma(n)}$, the conclusion is a partial sum of $\sum_{n\geq N_0}a_n$. Once we take absolute value on both sides, it is obvious that the absolute value of a partial sum of series is not larger than a partial sum of the series with absolute items, is not larger than a whole sum of the series with absolute items. – zyynankai Jun 06 '22 at 16:02